Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

The relative uncertainly in the period of a satellite orbiting around the earth is 10^{-2}. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :

A

10^{$$-$$2}

B

2 $$ \times $$ 10^{$$-$$2}

C

3 $$ \times $$ 10^{$$-$$2}

D

6 $$ \times $$ 10^{$$-$$2}

From kepler's law,

T = 2$$\pi $$ $$\sqrt {{{{r^3}} \over {GM}}} $$

$$ \Rightarrow $$$$\,\,\,$$T^{2} = $${{4{\pi ^2}} \over {GM}}{r^3}$$

$$ \Rightarrow $$$$\,\,\,$$ M = $${{4{\pi ^2}} \over G} \times {{{r^3}} \over {{T^2}}}$$

$$\therefore\,\,\,$$ $${{\Delta M} \over M}$$ = 2 $${{\Delta T} \over T}$$ + 3 $${{\Delta r} \over r}$$

as $${{\Delta r} \over r} \simeq 0$$

$$\therefore\,\,\,$$ $$\left| {{{\Delta M} \over M}} \right|$$ = 2 $${{\Delta T} \over T}$$ = 2 $$ \times $$ 10^{$$-$$2}

T = 2$$\pi $$ $$\sqrt {{{{r^3}} \over {GM}}} $$

$$ \Rightarrow $$$$\,\,\,$$T

$$ \Rightarrow $$$$\,\,\,$$ M = $${{4{\pi ^2}} \over G} \times {{{r^3}} \over {{T^2}}}$$

$$\therefore\,\,\,$$ $${{\Delta M} \over M}$$ = 2 $${{\Delta T} \over T}$$ + 3 $${{\Delta r} \over r}$$

as $${{\Delta r} \over r} \simeq 0$$

$$\therefore\,\,\,$$ $$\left| {{{\Delta M} \over M}} \right|$$ = 2 $${{\Delta T} \over T}$$ = 2 $$ \times $$ 10

2

A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is :

A

2.0 %

B

2.5 %

C

1.0 %

D

0.5 %

We know,

$$R = {{\rho l} \over A}$$

and Volume (V) = A$$l$$

$$ \Rightarrow $$ A $$=$$ $${V \over l}$$

$$ \therefore $$ $$R = {{\rho {l^2}} \over v}$$

$$ \therefore $$ $${{\Delta R} \over R} = 2{{\Delta l} \over l}$$

$$=$$ 2 $$ \times $$ 0.5

$$=$$ 1%

$$R = {{\rho l} \over A}$$

and Volume (V) = A$$l$$

$$ \Rightarrow $$ A $$=$$ $${V \over l}$$

$$ \therefore $$ $$R = {{\rho {l^2}} \over v}$$

$$ \therefore $$ $${{\Delta R} \over R} = 2{{\Delta l} \over l}$$

$$=$$ 2 $$ \times $$ 0.5

$$=$$ 1%

3

Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :

A

$$\sqrt {{{h{c^5}} \over G}} $$

B

$$\sqrt {{{{c^3}} \over {Gh}}} $$

C

$$\sqrt {{{Gh} \over {{c^5}}}} $$

D

$$\sqrt {{{Gh} \over {{c^3}}}} $$

Let t $$ \propto $$ G^{x} h^{y} c^{z}

$$ \therefore $$ [t] = [G]^{x} [h]^{y} [c]^{z} . . . . . (1)

We know,

F = $${{G{M^2}} \over {{R^2}}}$$

$$ \Rightarrow $$ G = $${{F{R^2}} \over {{M^2}}}$$

$$ \therefore $$ [G] = $${{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ {{M^2}} \right]}}$$

[G] = $$[{M^{ - 1}}{L^3}{T^{ - 2}}]$$

Also,

E = hf

$$ \therefore $$ [h] = $${{[E]} \over {[F]}}$$

= $${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {[{T^{ - 1}}]}}$$

= $$\left[ {M{L^2}{T^{ - 1}}} \right]$$

[C] = $$\left[ {{M^o}L\,{T^{ - 1}}} \right]$$

From equation (1) we get,

$$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} \right]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} \right]^y}{\left[ {{M^o}L{T^{ - 1}}} \right]^z}$$

$$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} \right]$$

By comparing the power of M, L, T

$$-$$ x + y = 0

$$ \Rightarrow $$ x = y

3x + 2y + z = 0

$$ \Rightarrow $$ 5x + z = 0 . . . . . (2)

$$-$$ 2x $$-$$ y $$-$$ z = 1

$$ \Rightarrow $$ $$-$$ 3x $$-$$ z = 1 . . . . (3)

By solving (2) and (3), we get,

x = $${1 \over 2}$$ = y and z = $$-$$ $${5 \over 2}$$

$$ \therefore $$ t $$ \propto $$ $$\sqrt {{{Gh} \over {{C^5}}}} $$

$$ \therefore $$ [t] = [G]

We know,

F = $${{G{M^2}} \over {{R^2}}}$$

$$ \Rightarrow $$ G = $${{F{R^2}} \over {{M^2}}}$$

$$ \therefore $$ [G] = $${{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ {{M^2}} \right]}}$$

[G] = $$[{M^{ - 1}}{L^3}{T^{ - 2}}]$$

Also,

E = hf

$$ \therefore $$ [h] = $${{[E]} \over {[F]}}$$

= $${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {[{T^{ - 1}}]}}$$

= $$\left[ {M{L^2}{T^{ - 1}}} \right]$$

[C] = $$\left[ {{M^o}L\,{T^{ - 1}}} \right]$$

From equation (1) we get,

$$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} \right]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} \right]^y}{\left[ {{M^o}L{T^{ - 1}}} \right]^z}$$

$$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} \right]$$

By comparing the power of M, L, T

$$-$$ x + y = 0

$$ \Rightarrow $$ x = y

3x + 2y + z = 0

$$ \Rightarrow $$ 5x + z = 0 . . . . . (2)

$$-$$ 2x $$-$$ y $$-$$ z = 1

$$ \Rightarrow $$ $$-$$ 3x $$-$$ z = 1 . . . . (3)

By solving (2) and (3), we get,

x = $${1 \over 2}$$ = y and z = $$-$$ $${5 \over 2}$$

$$ \therefore $$ t $$ \propto $$ $$\sqrt {{{Gh} \over {{C^5}}}} $$

4

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :

A

5.755 mm

B

5.950 mm

C

5.725 mm

D

5.740 mm

We know,

Least count (LC) = $${{Pitch} \over {no.\,of\,divisions}}$$

$$ \therefore $$ LC = $${{0.5} \over {100}}$$

= 0.5 $$ \times $$ 10^{$$-$$2} mm

Reading = MSR + CSR $$-$$ positive error

Given, Main scale reading (MSR) = 5.5 mm

Circular scale reading (CSR)

= 48 $$ \times $$ 0.5 $$ \times $$ 10^{$$-$$2} mm

= 0.24

As zero of its circular scale lines 3 division below the mean line, it means error is position error.

$$ \therefore $$ positive error

= 3 $$ \times $$ 0.5 $$ \times $$ 10^{$$-$$2} mm

= 0.015 mm

$$ \therefore $$ Reading = 5.5 + 0.24 $$-$$ 0.015

= 5.725 mm

Least count (LC) = $${{Pitch} \over {no.\,of\,divisions}}$$

$$ \therefore $$ LC = $${{0.5} \over {100}}$$

= 0.5 $$ \times $$ 10

Reading = MSR + CSR $$-$$ positive error

Given, Main scale reading (MSR) = 5.5 mm

Circular scale reading (CSR)

= 48 $$ \times $$ 0.5 $$ \times $$ 10

= 0.24

As zero of its circular scale lines 3 division below the mean line, it means error is position error.

$$ \therefore $$ positive error

= 3 $$ \times $$ 0.5 $$ \times $$ 10

= 0.015 mm

$$ \therefore $$ Reading = 5.5 + 0.24 $$-$$ 0.015

= 5.725 mm

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Units & Measurements *keyboard_arrow_right*

Motion *keyboard_arrow_right*

Laws of Motion *keyboard_arrow_right*

Work Power & Energy *keyboard_arrow_right*

Simple Harmonic Motion *keyboard_arrow_right*

Impulse & Momentum *keyboard_arrow_right*

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Gravitation *keyboard_arrow_right*

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Heat and Thermodynamics *keyboard_arrow_right*

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Vector Algebra *keyboard_arrow_right*

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Alternating Current and Electromagnetic Induction *keyboard_arrow_right*

Ray & Wave Optics *keyboard_arrow_right*

Atoms and Nuclei *keyboard_arrow_right*

Electronic Devices *keyboard_arrow_right*

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Practical Physics *keyboard_arrow_right*

Dual Nature of Radiation *keyboard_arrow_right*